Mathematics: Assume the time required to complete a product is normally distributed with a mean 3.2 hours and standard deviation .4 hours. How long should it take to complete a random unit in order to be in the top 10% (right tail) of the time distribution?

X/7 = 18/21Lets Reduce:21/7=3 so we know for every 3 in 18/21 we have one in x/7This means 6/7=18/21Check:18/6=321/7=3...

Assume the time required to complete a product is

adreyan3479

29.01.2022

3.712 hours or more

Step-by-step explanation:

Let X be the random variable that denotes the time required to complete a product.

X is normally distributed.

$X\sim N(\mu=3.2\text{ hours},\ \sigma=0.4\text{ hours} )$

Let x be the times it takes to complete a random unit in order to be in the top 10% (right tail) of the time distribution.

Then, $P(\dfrac{X-\mu}{\sigma}\dfrac{x-\mu}{\sigma})=0.10$

$P(z\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}]$

As, P(z1.28)=0.10 [By z-table]

Then,

$\dfrac{x-3.2}{0.4}=1.28\\\\\Rightarrow\ x=0.4\times1.28+3.2\\\\\Rightarrow\ x=3.712$

So, it will take 3.712 hours or more to complete a random unit in order to be in the top 10% (right tail) of the time distribution.

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